Question

A researcher is interested in determining the average number of years employees of a company stay with the company. If past information shows a standard deviation of 19 months, what size sample should be taken so that at 95% confidence, the margin of error will be 4 months or less?

Answer 1

z - z-score

e - margin of error

p - standard deviation

For 95% confidence z-score = 1.96

$\\n = \frac{z^2\times p^2}{e^2} = \frac{1.96^2\times19^2}{4^2}=$ 87

Answer: 87.